How Fast Is a Race Car After Traveling 200 Meters?

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SUMMARY

The discussion centers on calculating the speed of a race car after it has traveled 200 meters, starting from rest and accelerating uniformly at 4.90 m/s². The correct kinematic equation used is v² = v₀² + 2a(X - X₀), leading to a final speed of 44.3 m/s after substituting the values. Initial incorrect calculations led to a speed of 31.3 m/s, which was clarified through peer assistance. The importance of using the correct kinematic formula and constants is emphasized for accurate results.

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  • Ability to manipulate square roots and units in physics calculations
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Homework Statement


A race car starting from rest accelerates uniformly at a rate of 4.90m/s^{2} . What is the car's speed after it has traveled 200m?

Homework Equations



Δv = \frac{Change in position}{change in time}

The Attempt at a Solution


v=\sqrt{980m^2/s^2}

=31.3m/s
 
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vaironl said:

Homework Statement


A race car starting from rest accelerates uniformly at a rate of 4.90m/s^{2} . What is the car's speed after it has traveled 200m?

Homework Equations



Δv = \frac{Change in position}{change in time}

The Attempt at a Solution


v=\sqrt{980m^2/s^2}

=31.3m/s
Show your work so that we can help you see where your error is.

How did you get v=\sqrt{980m^2/s^2}\,?
 
What kinematic formula are you using? It looks like you've missed out a constant.
 
SammyS said:
Show your work so that we can help you see where your error is.

How did you get v=\sqrt{980m^2/s^2}\,?

gneill said:
What kinematic formula are you using? It looks like you've missed out a constant.

Well I'm required to find the speed after 200 meters, but I was never given any time.

So I assume... Which is not a good thing to do, multiplying the acceleration * the distanced travel would give me the speed.

But I really can't remember the actual equation.
 
vaironl said:
Well I'm required to find the speed after 200 meters, but I was never given any time.

So I assume... Which is not a good thing to do, multiplying the acceleration * the distanced travel would give me the speed.

But I really can't remember the actual equation.

If you can't remember you should check your notes or textbook! Your formula is close to being a valid kinematic expression,... but it's missing a constant. Do you have a list of the common kinematic expressions?
 
gneill said:
If you can't remember you should check your notes or textbook! Your formula is close to being a valid kinematic expression,... but it's missing a constant. Do you have a list of the common kinematic expressions?

I have the textbook, and I did have an example sheet but my friend borrowed to check her work. I found this expression which seems to satisfy my problem, v2 = vo2 + 2a(X - Xo)

Therefore V2= 0m/s + 2(4.90m/s^2)(200m)

v2= 9.8m/s^2(200m)
v2= 1960m^2/s^2
v=\sqrt{1960m^2/s^2}
v= 44.3m/s
 
vaironl said:
I have the textbook, and I did have an example sheet but my friend borrowed to check her work. I found this expression which seems to satisfy my problem, v2 = vo2 + 2a(X - Xo)

Therefore V2= 0m/s + 2(4.90m/s^2)(200m)

v2= 9.8m/s^2(200m)
v2= 1960m^2/s^2
v=\sqrt{1960m^2/s^2}
v= 44.3m/s

That looks better :smile:
 

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